Question

The value of $\underset{x\to \infty }{lim}[\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}]$ is

• A. $\frac{1}{2}$
• B. $1$
• C. $0$
• D. None of these

Answer 1

The correct option is A $\frac{1}{2}$

${lim}_{x⟶\infty }[\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}]$

divide and multiply by $\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}$,

We get,

$\Rightarrow {lim}_{x⟶\infty }\left[\frac{x+\sqrt{x+\sqrt{x}}-x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\right]$

(After canceling)

$\Rightarrow {lim}_{x⟶\infty }\left[\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\right]$

$\Rightarrow divideandmultiplyby\sqrt{x}$

$\Rightarrow {lim}_{x⟶\infty }\left[\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x^{3/2}}}}+1}\right]$

$putx$ tends to $\infty$

$\Rightarrow \frac{\sqrt{1+0}}{\sqrt{1+\sqrt{0+0}}+1}$

$=\frac{1}{2}$